What is inverse oflaplace (3s)/((s-1)(s^2-4)) ?

The answer to inverse oflaplace (3s)/((s-1)(s^2-4)) is -e^t-1/2 e^{-2t}+3/2 e^{2t}

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inverse oflaplace (3s)/((s-1)(s^2-4))

inverse laplace

\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}

- e^{t} - \frac{1}{2} e^{- 2 t} + \frac{3}{2} e^{2 t}

Solution steps

L^{- 1} {\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}}

Take the partial fraction of

\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )} : - \frac{1}{s - 1} - \frac{1}{2 ( s + 2 )} + \frac{3}{2 ( s - 2 )}

= L^{- 1} {- \frac{1}{s - 1} - \frac{1}{2 ( s + 2 )} + \frac{3}{2 ( s - 2 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{1}{s - 1}} - L^{- 1} {\frac{1}{2 ( s + 2 )}} + L^{- 1} {\frac{3}{2 ( s - 2 )}}

L^{- 1} {\frac{1}{s - 1}} : e^{t}

L^{- 1} {\frac{1}{2 ( s + 2 )}} : \frac{1}{2} e^{- 2 t}

L^{- 1} {\frac{3}{2 ( s - 2 )}} : \frac{3}{2} e^{2 t}

= - e^{t} - \frac{1}{2} e^{- 2 t} + \frac{3}{2} e^{2 t}

\int \frac{( 6 x - 7 )}{3 x ^{2} - 7 x + 11} d x

(2, - 3), (- 1, 15)

\frac{\partial}{\partial x} (x^{5} + 3 x^{3} y^{2} + 3 x y^{4})

x \to 1 lim (\frac{x ^{2}}{x - 1})

\frac{\partial}{\partial y} (x y^{2} + x^{2} y)

What is inverse oflaplace (3s)/((s-1)(s^2-4)) ?
The answer to inverse oflaplace (3s)/((s-1)(s^2-4)) is -e^t-1/2 e^{-2t}+3/2 e^{2t}